Reduction of order in ODE $$x^2 y'' - 3 x y' + 4y = 0$$
where $y_1=x^2 \ln (x)$ is a solution. I need to use the given solution $y_1$ to find a second order linearly independent solution.
So $y = y_1 v$ and $y =v x^2 \ln (x)$. I know I must find the first and second derivatives of y and plug them back into the equation and then make a substitution $w=v'$, $w'=v''$ but I an not sure how to work out the first and second derivatives and them plug them in to cancel down 
 A: Assume we are solving the ODE away from $x=0$. the second solution is of the form $y_{2}(x)=x^{2}\log(x)v(x)$.
Note that 
$$
y_{2}^{\prime}(x)=x\left(x\log(x)v^{\prime}(x)+v(x)+2\log(x)v(x)\right)
$$
 and 
$$
y_{2}^{\prime\prime}(x)=x\left(x\log(x)v^{\prime\prime}(x)+\left(4\log(x)+2\right)v^{\prime}(x)\right)+v(x)\left(2\log(x)+3\right).
$$
Plugging these into the original equation and simplying,
$$
x^{3}\left(x\log(x)v^{\prime\prime}(x)+\left(\log(x)+2\right)v^{\prime}(x)\right)=0.
$$
Since we are solving away from $x=0$, we can divide both sides by
$x^{3}$. Now, let $w=v^{\prime}$ so that
$$
x\log(x)w^{\prime}(x)+\left(\log(x)+2\right)w(x)=0.
$$
This is a separable equation with solution $w(x)=c_{1}/(x\log^{2}(x))$.
Integrating, we get $v(x)=-c_{1}/\log(x)+c_{2}$. Therefore, 
$$
y_{2}(x)=x^{2}\log(x)v(x)=-c_{1}x^{2}+c_{2}.
$$
Since we are only looking for a particular solution, set $c_{1}=-1$
and $c_{2}=0$ to get $y_{2}(x)=x^{2}$.

Now, a linear combination of solutions is also a solution since the ODE itself is linear (check). In your case,
$y(x)=c_{1}x^{2}+c_{2}x^{2}\log(x)$ is a solution. To determine $c_{1}$
and $c_{2}$, you need more information (e.g. the value of $y(x)$
and $y^{\prime}(x)$ at some $x=x_{0}>0$).
Let's say we were told $y(x_{0})=y_{0}$
and $y^{\prime}(x_{0})=y_{0}^{\prime}$. Then, differentiating the
solution yields $y^{\prime}(x)=x(2c_{1}+c_{2})+2c_{2}x\log x$. Therefore,
we get the system of equations
\begin{align*}
y(x_{0}) & =c_{1}x_{0}^{2}+c_{2}x_{0}^{2}\log(x_{0});\\
y^{\prime}(x_{0}) & =x_{0}(2c_{1}+c_{2})+2c_{2}x_{0}\log x_{0}.
\end{align*}
That's two equations, two unknowns. You can now solve for $c_{1}$
and $c_{2}$. 
